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Quadratic Algebra Fractions

  1. Sep 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Hello, I just found a question, and having attempted it many times I get different answers, probably due to my messy working, however I have just tried it twice again and got the same answer, just checking with you guys to see if you think it is correct.


    Simplify fully


    My answer is 1/(x+4)

    Hope it's right!

  2. jcsd
  3. Sep 18, 2007 #2
    Is this the fraction [tex]\frac{2}{x-1} + \frac{x-11}{x^2+3x-4}[/tex]? If so, you have missed a factor of 3.
  4. Sep 18, 2007 #3


    What do you mean?

    yes the question is as you stated
  5. Sep 18, 2007 #4
    I started out solving the question by getting lowest common denominator.. i.e

    (x-1)(x+4)+(x-1) was my bottom line.. and then i carried on.. so which factor did I miss?
  6. Sep 18, 2007 #5
    What did you do after that?
  7. Sep 18, 2007 #6

    Ok here is the working..


    Simplifies into..


    which then is


    which then is


    which then turns out to be

    (x-1)(x-1)/(x-1)(x+4)(x-1) = 1/(x+4)

    I think..
  8. Sep 18, 2007 #7
    Hmm...I'm not sure how you arrived at that, but

    [tex]\frac{2}{x-1} + \frac{x-11}{x^2+3x-4}[/tex] = [tex]\frac{2(x+4) +(x-11)}{(x-1)(x+4)} [/tex]

    Think of it as multiplying the numerator and denominator of the first term by (x+4) and simplifying.
  9. Sep 18, 2007 #8
    ahhhhh i get you, i understand, didnt spot that (x-1) was also factor or the other quadratic.. but 1 thing..

    the way I got to my answer is taking the LCM by multiplying the two bottom terms, then working from there.. theoretically it should have come out right, i can't see why it didn't
  10. Sep 18, 2007 #9
    The LCM of (x-1) and (x-1)(x+4) is (x-1)(x+4). Therefore the denominator becomes (x-1)(x+4), and you multiply the 2 by (x+4) and the (x-11) by 1.
  11. Sep 18, 2007 #10
    yeah I get that, but couldn't (x-1)(x-1)(x+4) be used too as its still a multiple of both
  12. Sep 18, 2007 #11
    We're talking about least common multiple.
  13. Sep 18, 2007 #12
    I thought when adding fractions you just needed a common multiple, not the lowest? or am I wrong?
  14. Sep 18, 2007 #13
    You said you took the LCM. What do you think LCM stands for? :wink:

    Here's a numerical example: (1/2) + (1/4) = 0.50 + 0.25 = 0.75

    The least common multiple of 2 and 4 is 4, since 2x2 = 4x1 = 4.

    (1/2) + (1/4) = ((1x2) + (1x1))/4 = 3/4 = 0.75
    Last edited: Sep 18, 2007
  15. Sep 18, 2007 #14
    Yeah, I thought I took what I thought was the LCM, I was wrong, but what i'm asking is that shouldn't it have worked as the denominator I gave was still a multiple of both
  16. Sep 18, 2007 #15
    Ah. What you essentially did was multiply the numerator and denominator by (x-1). So the answer should be the same as the correct one.

    But from going from 3x^2-6x+3/(x-1)(x+4)(x-1) to x^2-2x+1/(x-1)(x+4)(x-1), you forgot the factor of three in the numerator, just as I said in my first post. :biggrin:
  17. Sep 18, 2007 #16
    But doesn't 3x^2-6x+3 = x^2-2x+1? As I took out the factor of 3?
  18. Sep 18, 2007 #17
    But where does the 3 go?

    What you're saying is the same as stating 3 = 1. Does that make sense?

    3x^2-6x+3 is equal to thrice x^2-2x+1.

    Since 3 is common to all terms, I can write it as 3(x^2-2x+1). Multiply each term within the brackets by 3 and you'll get back to old expression.
  19. Sep 18, 2007 #18
    hmm ok.. kinda wierd, since i thought dividing everything by three should keep it the same
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